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**iLikeMaths** for which value of the constant $\displaystyle K$ is the function $\displaystyle h(x)=\left\{\begin{array}{cc}\frac{e^x -e^a}{x-a},&\mbox{ if }

x\ne a\\K, & \mbox{ if } x=a\end{array}\right.$

continuous at $\displaystyle x=a$. justify your answer.

Answer

$\displaystyle \lim_{x \rightarrow a} h(x) = \lim_{x \rightarrow a} \frac{e^x - e^a}{x-a}$

$\displaystyle h(a) = K$ and the other limit is $\displaystyle e^a \log(e)$

so is $\displaystyle K = e^a \log(e)$?, is this a correct way of answering the question